27 28 28 triangle

Acute isosceles triangle.

Sides: a = 27   b = 28   c = 28

Area: T = 331.1633007445
Perimeter: p = 83
Semiperimeter: s = 41.5

Angle ∠ A = α = 57.65108983208° = 57°39'3″ = 1.00661979924 rad
Angle ∠ B = β = 61.17545508396° = 61°10'28″ = 1.06876973306 rad
Angle ∠ C = γ = 61.17545508396° = 61°10'28″ = 1.06876973306 rad

Height: ha = 24.53105931441
Height: hb = 23.65545005318
Height: hc = 23.65545005318

Median: ma = 24.53105931441
Median: mb = 23.67548812035
Median: mc = 23.67548812035

Inradius: r = 7.98798315047
Circumradius: R = 15.98800457208

Vertex coordinates: A[28; 0] B[0; 0] C[13.01878571429; 23.65545005318]
Centroid: CG[13.67326190476; 7.88548335106]
Coordinates of the circumscribed circle: U[14; 7.70546649011]
Coordinates of the inscribed circle: I[13.5; 7.98798315047]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 122.3499101679° = 122°20'57″ = 1.00661979924 rad
∠ B' = β' = 118.825544916° = 118°49'32″ = 1.06876973306 rad
∠ C' = γ' = 118.825544916° = 118°49'32″ = 1.06876973306 rad

Calculate another triangle




How did we calculate this triangle?

Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.

a = 27 ; ; b = 28 ; ; c = 28 ; ;

1. The triangle circumference is the sum of the lengths of its three sides

p = a+b+c = 27+28+28 = 83 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 83 }{ 2 } = 41.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 41.5 * (41.5-27)(41.5-28)(41.5-28) } ; ; T = sqrt{ 109668.94 } = 331.16 ; ;

4. Calculate the heights of the triangle from its area.

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 331.16 }{ 27 } = 24.53 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 331.16 }{ 28 } = 23.65 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 331.16 }{ 28 } = 23.65 ; ;

5. Calculation of the inner angles of the triangle using a Law of Cosines

a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 27**2-28**2-28**2 }{ 2 * 28 * 28 } ) = 57° 39'3" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 28**2-27**2-28**2 }{ 2 * 27 * 28 } ) = 61° 10'28" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-27**2-28**2 }{ 2 * 28 * 27 } ) = 61° 10'28" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 331.16 }{ 41.5 } = 7.98 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 27 }{ 2 * sin 57° 39'3" } = 15.98 ; ;




Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.